Let $X$ be a submartingale and $\sigma$ and $\tau$ two bounded stopping, then $E[X_{\tau}\vert \mathcal{F}_\sigma] \geq X_{\sigma \wedge \tau}$.

Question

Let $X$ be a submartingale and $\sigma$ and $\tau$ two bounded stopping, then $E[X_{\tau}\vert \mathcal{F}_\sigma] \geq X_{\sigma \wedge \tau}$.

How to use the fact that $E[X_\tau\vert \mathcal{F}_\sigma] \geq X_\sigma$ if $\tau > \sigma$ to prove this statement?

Answer 1

Write $X_\tau = X_{\sigma\vee \tau}1[{\tau > \sigma}] + X_{\sigma\wedge \tau} 1[\tau \leq \sigma]$ and ask yourself what parts are $\mathcal{F}_\sigma$-measurable.

Answer 2

The stopped submartingale $X^\tau_t:=X_{t\wedge\tau}$ is a submartingale. In particular, if $\nu$ is a bounded stopping time with $\nu\ge\sigma$, then $E[X^\tau_\nu\mid\mathcal F_\sigma]\ge X^\tau_\sigma$. What's a suitable choice for $\nu$?